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Solutions for Chapter 1
Solutions for exercises in section 1. 2
1.2.1. (1,0,0)
1.2.2. (1,2,3)
1.2.3. (1,0,−1)
1.2.4. (−1/2,1/2,0,1)
1.2.5.
2 −4 3
4 −7 4
5 −8 4
1.2.6. Every row operation is reversible. In particular the “inverse” of any row operation
is again a row operation of the same type.
1.2.7. π
2, π,0
1.2.8. The third equation in the triangularized form is 0x3 = 1, which is impossible
to solve.
1.2.9. The third equation in the triangularized form is 0x3 = 0, and all numbers are
solutions. This means that you can start the back substitution with any value whatsoever and consequently produce infinitely many solutions for the system.
1.2.10. α=−3, β=112, and γ=−3
2
1.2.11. (a) If xi= the number initially in chamber #i, then
.4x1+ 0x2+ 0x3+.2x4= 12
0x1+.4x2+.3x3+.2x4= 25
0x1+.3x2+.4x3+.2x4= 26
.6x1+.3x2+.3x3+.4x4= 37
and the solution is x1= 10, x2= 20, x3= 30, and x4= 40.
(b) 16, 22, 22, 40
1.2.12. To interchange rows i and j, perform the following sequence of Type II and
Type III operations.
Rj←Rj+Ri (replace rowj by the sum of rowj andi)
Ri←Ri−Rj (replace rowi by the difference of rowiandj)
Rj←Rj+Ri (replace rowj by the sum of rowj andi)
Ri← −Ri (replace rowiby its negative)
1.2.13. (a) This has the effect of interchanging the order of the unknowns—xj and
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solution to the old system except that the solution for the jth unknown of the
new system is ˆxj =α1xj. This has the effect of “changing the units” of the jth
unknown. (c) The solution to the new system is the same as the solution for the old system except that the solution for the kth unknown in the new system
is ˆxk=xk−αxj.
1.2.14. hij =i+1j−1
1.2.16. If x=
x1
x2
.. .
xm
and y=
y1
y2
.. .
ym
are two different solutions, then
z= x+y
2 =
x1+y1
2
x2+y2
2
.. .
xm+ym
2
is a third solution different from both x and y.
Solutions for exercises in section 1. 3
1.3.1. (1,0,−1)
1.3.2. (2,−1,0,0)
1.3.3.
1 1 11 2 2 1 2 3
Solutions for exercises in section 1. 4
1.4.2. Use y′(tk) =y′k≈
yk+1−yk−1
2h and y
′′(t
k) =yk′′≈
yk−1−2yk+yk+1
h2 to write
f(tk) =fk=y′′k−yk′ ≈
2yk−1−4yk+ 2yk+1
2h2 −
hyk+1−hyk−1
2h2 , k= 1,2, . . . , n,
with y0=yn+1= 0. These discrete approximations form the tridiagonal system
−4 2−h
2 +h −4 2−h
. .. . .. . ..
2 +h −4 2−h
2 +h −4
y1
y2
.. .
yn−1
yn
= 2h
2
f1
f2
.. .
fn−1
fn
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Solutions for exercises in section 1. 5
1.5.1. (a) (0,−1) (c) (1,−1) (e) 1
1.001, − 1 1.001
1.5.2. (a) (0,1) (b) (2,1) (c) (2,1) (d) 1.00012 ,11..00030001
1.5.3. Without PP: (1.01, 1.03) With PP: (1, 1) Exact: (1, 1)
1.5.4. (a)
.5001 ..500333 ..333250 ..333333
.333 .250 .200 .200 −→
10 ..500083 ..333083 ..333166 0 .083 .089 .089
−→
10 ..500083 ..333083 ..333166 0 0 .006 −.077
z=−.077/.006 =−12.8,
y= (.166−.083z)/.083 = 14.8, x=.333−(.5y+.333z) =−2.81
(b)
1 .500 .333 .333
.500 .333 .250 .333
.333 .250 .200 .200 −→
1 .500 .333 .333 1 .666 .500 .666 1 .751 .601 .601
−→
10 ..500166 ..333167 ..333333 0 .251 .268 .268
−→
10 ..500251 ..333268 ..333268 0 .166 .167 .333
−→
10 ..500251 ..333268 ..333268
0 0 −.01 .156
z=−.156/.01 =−15.6,
y= (.268−.268z)/.251 = 17.7, x=.333−(.5y+.333z) =−3.33
(c)
.5001 ..500333 ..333250 ..333333
.333 .250 .200 .200 −→
11 ..500666 ..333500 ..333666 1 .751 .601 .601
−→
1 .500 .333 .333 0 .166 .167 .333 0 .251 .268 .268
−→
1 .500 .333 .333
0 .994 1 1.99
0 .937 1 1
−→
10 ..500994 .3331 .1333.99 0 0 .057 −.880
z=−.88/.057 =−15.4,
y= (1.99−z)/.994 = 17.5, x=.333−(.5y+.333z) =−3.29 (d) x=−3, y= 16, z=−14
1.5.5. (a)
.0055x+.095y+ 960z= 5000
.0011x+. 01y+ 112z= 600
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(b) 3-digit solution = (55,900 lbs. silica, 8,600 lbs. iron, 4.04 lbs. gold).
Exact solution (to 10 digits) = (56,753.68899,8,626.560726,4.029511918). The relative error (rounded to 3 digits) is er= 1.49×10−2.
(c) Let u=x/2000, v=y/1000, and w= 12z to obtain the system
11u+ 95v+ 80w= 5000 2.2u+ 10v+ 9.33w= 600 18.6u+ 25v+ 46.7w= 3000.
(d) 3-digit solution = (28.5 tons silica, 8.85 half-tons iron, 48.1 troy oz. gold).
Exact solution (to 10 digits) = (28.82648317,8.859282804,48.01596023). The relative error (rounded to 3 digits) is er = 5.95×10−3. So, partial pivoting
applied to the column-scaled system yields higher relative accuracy than partial pivoting applied to the unscaled system.
1.5.6. (a) (−8.1,−6.09) = 3-digit solution with partial pivoting but no scaling.
(b) No! Scaled partial pivoting produces the exact solution—the same as with complete pivoting.
1.5.7. (a) 2n−1 (b) 2
(c) This is a famous example that shows that there are indeed cases where par-tial pivoting will fail due to the large growth of some elements during elimination, but complete pivoting will be successful because all elements remain relatively small and of the same order of magnitude.
1.5.8. Use the fact that with partial pivoting no multiplier can exceed 1 together with
the triangle inequality |α+β| ≤ |α|+|β| and proceed inductively.
Solutions for exercises in section 1. 6
1.6.1. (a) There are no 5-digit solutions. (b) This doesn’t help—there are now infinitely
many 5-digit solutions. (c) 6-digit solution = (1.23964,−1.3) and exact solution = (1,−1) (d) r1=r2= 0 (e) r1=−10−6 and r2= 10−7 (f) Even if computed
residuals are 0, you can’t be sure you have the exact solution.
1.6.2. (a) (1,−1.0015) (b) Ill-conditioning guarantees that the solution will be very
sensitive tosomesmall perturbation but not necessarily toevery small perturba-tion. It is usually difficult to determine beforehand those perturbations for which an ill-conditioned system will not be sensitive, so one is forced to be pessimistic whenever ill-conditioning is suspected.
1.6.3. (a) m1(5) = m2(5) =−1.2519, m1(6) = −1.25187, and m2(6) = −1.25188
(c) An optimally well-conditioned system represents orthogonal (i.e., perpen-dicular) lines, planes, etc.
1.6.4. They rank as (b) = Almost optimally conditioned. (a) = Moderately
well-conditioned. (c) = Badly ill-well-conditioned.
1.6.5. Original solution = (1,1,1). Perturbed solution = (−238,490,−266). System
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Solutions for Chapter 2
Solutions for exercises in section 2. 1
2.1.1. (a)
10 22 31 30
0 0 0 3
is one possible answer. Rank = 3 and the basic columns
are {A∗1,A∗2,A∗4}. (b)
1 2 3
0 2 2
0 0 −8
0 0 0
0 0 0
is one possible answer. Rank = 3 and
every column in A is basic.
(c)
2 1 1 3 0 4 1
0 0 2 −2 1 −3 3
0 0 0 0 −1 3 −1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
is one possible answer. The rank is 3, and
the basic columns are {A∗1,A∗3,A∗5}.
2.1.2. (c) and (d) are in row echelon form.
2.1.3. (a) Since any row or column can contain at most one pivot, the number of pivots
cannot exceed the number of rows nor the number of columns. (b) A zero row cannot contain a pivot. (c) If one row is a multiple of another, then one of them can be annihilated by the other to produce a zero row. Now the result of the previous part applies. (d) One row can be annihilated by the associated combination of row operations. (e) If a column is zero, then there are fewer than
n basic columns because each basic column must contain a pivot.
2.1.4. (a) rank(A) = 3 (b) 3-digit rank(A) = 2 (c) With PP, 3-digit rank(A) = 3
2.1.5. 15
2.1.6. (a) No, consider the form
∗ ∗ ∗0 0 0 ∗0
0 0 0 ∗
(b) Yes—in fact, E is a row
echelon form obtainable from A.
Solutions for exercises in section 2. 2
2.2.1. (a)
1 00 1 212 00
0 0 0 1
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(b)
1 1
2 0 2 0 2 0
0 0 1 −1 0 0 1
0 0 0 0 1 −3 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
and
A∗2= 12A∗1, A∗4= 2A∗1−A∗3, A∗6= 2A∗1−3A∗5, A∗7=A∗3+A∗5
2.2.2. No.
2.2.3. The same would have to hold in EA, and there you can see that this means not
all columns can be basic. Remember, rank(A) = number of basic columns.
2.2.4. (a)
10 01 00
0 0 1
(b)
10 01 −12
0 0 0
A∗3 is almost a combination of A∗1
and A∗2. In particular, A∗3≈ −A∗1+ 2A∗2.
2.2.5. E∗1= 2E∗2−E∗3 and E∗2=12E∗1+12E∗3
Solutions for exercises in section 2. 3
2.3.1. (a), (b)—There is no need to do any arithmetic for this one because the
right-hand side is entirely zero so that you know (0,0,0) is automatically one solution. (d), (f)
2.3.3. It is always true that rank(A) ≤ rank[A|b] ≤ m. Since rank(A) = m, it
follows that rank[A|b] =rank(A).
2.3.4. Yes—Consistency implies that b and c are each combinations of the basic
columns in A. If b= βiA∗bi and c= γiA∗bi where the A∗bi’s are the
basic columns, then b+c= (βi+γi)A∗bi = ξiA∗bi, where ξi =βi+γi
so that b+c is also a combination of the basic columns in A.
2.3.5. Yes—because the 4×3 system α+βxi+γx2i =yi obtained by using the four
given points (xi, yi) is consistent.
2.3.6. The system is inconsistent using 5-digits but consistent when 6-digits are used.
2.3.7. If x, y, and z denote the number of pounds of the respective brands applied,
then the following constraints must be met.
total # units of phosphorous = 2x+ y+z= 10 total # units of potassium = 3x+ 3y = 9
total # units of nitrogen = 5x+ 4y+z= 19
Since this is a consistent system, the recommendation can be satisfied exactly. Of course, the solution tells how much of each brand to apply.
2.3.8. No—if one or more such rows were ever present, how could you possibly eliminate
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Solutions for exercises in section 2. 4
2.4.1. (a) x2
−2 1 0 0 +x4
−1 0 −1 1
(b) y
− 1 2 1 0
(c) x3
−1 −1 1 0 +x4
−1 1 0 1
(d) The trivial solution is the only solution.
2.4.2.
00
0 and
−11
2
0
2.4.3. x2
−2 1 0 0 0 +x4
−2 0 −1 1 0
2.4.4. rank(A) = 3
2.4.5. (a) 2—because the maximum rank is 4. (b) 5—because the minimum rank is
1.
2.4.6. Because r=rank(A)≤m < n =⇒ n−r >0.
2.4.7. There are many different correct answers. One approach is to answer the question
“What must EA look like?” The form of the general solution tells you that
rank(A) = 2 and that the first and third columns are basic. Consequently,
EA =
10 α0 01 βγ
0 0 0 0
so that x1 =−αx2−βx4 and x3 =−γx4 gives rise
to the general solution x2
−α 1 0 0 +x4
−β 0 −γ 1
. Therefore, α = 2, β = 3,
and γ = −2. Any matrix A obtained by performing row operations to EA
will be the coefficient matrix for a homogeneous system with the desired general solution.
2.4.8. If ixfihi is the general solution, then there must exist scalars αi and βi such
that c1 = iαihi and c2 = iβihi. Therefore, c1+c2 = i(αi+βi)hi,
and this shows that c1+c2 is the solution obtained when the free variables xfi
assume the values xfi=αi+βi.
Solutions for exercises in section 2. 5
2.5.1. (a) 1 0 2 0 +x2
−2 1 0 0 +x4
−1 0 −1 1 (b) 10
2 +y
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(c) 2 −1 0 0 +x3 −1 −1 1 0 +x4
−1 1 0 1 (d) −33
−1
2.5.2. From Example 2.5.1, the solutions of the linear equations are:
x1= 1−x3−2x4
x2= 1−x3
x3 is free
x4 is free
x5=−1
Substitute these into the two constraints to get x3 =±1 and x4 =±1. Thus
there are exactly four solutions:
−2 0 1 1 −1 , 2 0 1 −1 −1 , 0 2 −1 1 −1 , 4 2 −1 −1 −1
2.5.3. (a) {(3,0,4), (2,1,5), (1,2,6), (0,3,7)} See the solution to Exercise 2.3.7 for
the underlying system. (b) (3,0,4) costs $15 and is least expensive.
2.5.4. (a) Consistent for all α. (b) α= 3, in which case the solution is (1,−1,0).
(c) α= 3, in which case the general solution is 1 −1 0 +z
0 −3 2 1 . 2.5.5. No 2.5.6.
EA=
1 0 · · · 0 0 1 · · · 0
..
. ... . .. ... 0 0 · · · 1 0 0 · · · 0
..
. ... · · · ... 0 0 · · · 0
m×n
2.5.7. See the solution to Exercise 2.4.7.
2.5.8. (a)
−.39760
1 +y
−.79881
0
(b) There are no solutions in this case.
(c)
1.−439642.3 1
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Solutions for exercises in section 2. 6
2.6.1. (a) (1/575)(383,533,261,644,−150,−111)
2.6.2. (1/211)(179,452,36)
2.6.3. (18, 10)
2.6.4. (a) 4 (b) 6 (c) 7 loops but only 3 simple loops. (d) Show that
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Solutions for Chapter 3
Solutions for exercises in section 3. 2
3.2.1. (a) X=
0 1 2 3
(b) x=−1
2, y=−6, and z= 0
3.2.2. (a) Neither (b) Skew symmetric (c) Symmetric (d) Neither
3.2.3. The 3×3 zero matrix trivially satisfies all conditions, and it is the only
pos-sible answer for part (a). The only pospos-sible answers for (b) are real symmetric matrices. There are many nontrivial possibilities for (c).
3.2.4. A = AT and B = BT =
⇒ (A+B)T = AT +BT = A+B. Yes—the
skew-symmetric matrices are also closed under matrix addition.
3.2.5. (a) A=−AT =
⇒ aij=−aji. If i=j, then ajj=−ajj =⇒ ajj= 0.
(b) A=−A∗ =⇒ aij =−aji. If i=j, then ajj=−ajj. Write ajj=x+ iy
to see that ajj=−ajj =⇒ x+ iy=−x+ iy =⇒ x= 0 =⇒ ajj is pure
imaginary.
(c) B∗= (iA)∗=−iA∗=−iAT =−iAT =
−iA=−B.
3.2.6. (a) Let S=A+AT and K=A
−AT. Then ST =AT+AT T =AT+A=S.
Likewise, KT =AT −AT T =AT −A=−K.
(b) A=S
2+
K
2 is one such decomposition. To see it is unique, suppose A=X+
Y, where X=XT and Y=−YT. Thus, AT =XT+YT =X−Y =⇒ A+
AT = 2X, so that X = A+AT
2 =
S
2. A similar argument shows that Y =
A−AT
2 =
K
2.
3.2.7. (a) [(A+B)∗]ij = [A+B]ji = [A+B]ji = [A]ji+ [B]ji= [A∗]ij + [B∗]ij =
[A∗+B∗]
ij
(b) [(αA)∗]ij= [αA]ji= [¯αA]ji= ¯α[A]ji= ¯α[A∗]ij
3.2.8. k
1 −1 0 · · · 0 0
−1 2 −1 · · · 0 0 0 −1 2 · · · 0 0
..
. ... ... . .. ... ... 0 0 0 · · · 2 −1 0 0 0 · · · −1 1
Solutions for exercises in section 3. 3
3.3.1. Functions (b) and (f) are linear. For example, to check if (b) is linear, let
A=
a1
a2
and B =
b1
b2
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f(αA) =αf(A). Do so by writing
f(A+B) =f
a1+b1
a2+b2
=
a2+b2
a1+b1
=
a2
a1
+
b2
b1
=f(A) +f(B),
f(αA) =f
αa1
αa2
=
αa2
αa1
=α
a2
a1
=αf(A).
3.3.2. Write f(x) = ni=1ξixi. For all points x=
x1
x2
.. .
xn
and y=
y1
y2
.. .
yn
, and for
all scalars α, it is true that
f(αx+y) =
n
i=1
ξi(αxi+yi) = n
i=1
ξiαxi+ n
i=1
ξiyi
=α
n
i=1
ξixi+ n
i=1
ξiyi=αf(x) +f(y).
3.3.3. There are many possibilities. Two of the simplest and most common are Hooke’s
law for springs that says that F =kx (see Example 3.2.1) and Newton’s second law that says that F =ma (i.e., force = mass×acceleration).
3.3.4. They are all linear. To see that rotation is linear, use trigonometry to deduce
that if p=
x1
x2
, then f(p) =u=
u1
u2
, where
u1= (cosθ)x1−(sinθ)x2
u2= (sinθ)x1+ (cosθ)x2.
f is linear because this is a special case of Example 3.3.2. To see that reflection is linear, write p =
x1
x2
and f(p) =
x1
−x2
. Verification of linearity is straightforward. For the projection function, use the Pythagorean theorem to
conclude that if p=
x1
x2
, then f(p) =x1+x2
2
1 1
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Solutions for exercises in section 3. 4
3.4.1. Refer to the solution for Exercise 3.3.4. If Q, R, and P denote the matrices
associated with the rotation, reflection, and projection, respectively, then
Q=
cosθ −sinθ
sinθ cosθ
, R=
1 0
0 −1
, and P= 1
2 1 2 1 2
1 2
.
3.4.2. Refer to the solution for Exercise 3.4.1 and write
RQ=
1 0
0 −1
cosθ −sinθ
sinθ cosθ
=
cosθ −sinθ −sinθ −cosθ
.
If Q(x) is the rotation function and R(x) is the reflection function, then the composition is
RQ(x)=
(cosθ)x1−(sinθ)x2
−(sinθ)x1−(cosθ)x2
.
3.4.3. Refer to the solution for Exercise 3.4.1 and write
PQR=
a11x1+a12x2
a21x1+a22x2
cosθ −sinθ
sinθ cosθ
1 0
0 −1
= 1 2
cosθ+ sinθ sinθ−cosθ
cosθ+ sinθ sinθ−cosθ
.
Therefore, the composition of the three functions in the order asked for is
P
QR(x)
= 1 2
(cosθ+ sinθ)x1+ (sinθ−cosθ)x2
(cosθ+ sinθ)x1+ (sinθ−cosθ)x2
.
Solutions for exercises in section 3. 5
3.5.1. (a) AB =
1012 158
28 52
(b) BA does not exist (c) CB does not exist
(d) CTB= ( 10 31 ) (e) A2=
1316 −131 1912 36 −17 64
(f) B2 does not exist
(g) CTC = 14 (h) CCT =
1 2 32 4 6 3 6 9
(i) BBT =
58 16 288 17 17 28 58
(j) BTB=
10 23 23 69
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3.5.2. (a) A=
24 10 12
2 2 0
,x=
xx12
x3
,b=
103
−2
(b) s= −12
3
(c) b=A∗1−2A∗2+ 3A∗3
3.5.3. (a) EA=
AA12∗∗
3A1∗+A3∗
(b) AE= (A∗1+ 3A∗3 A∗2 A∗3)
3.5.4. (a) A∗j (b) Ai∗ (c) aij
3.5.5. Ax=Bx ∀ x =⇒ Aej =Bej ∀ ej =⇒ A∗j =B∗j ∀ j =⇒ A=B.
(The symbol ∀ is mathematical shorthand for the phrase “for all.”)
3.5.6. The limit is the zero matrix.
3.5.7. If A is m×p and B is p×n, write the product as
AB= (A∗1 A∗2 · · · A∗p)
B1∗
B2∗
.. .
Bp∗
=A∗1B1∗+A∗2B2∗+· · ·+A∗pBp∗
=
p
k=1
A∗kBk∗.
3.5.8. (a) [AB]ij =Ai∗B∗j = ( 0 · · · 0 aii · · · ain)
b1j
.. .
bjj
0 .. . 0
is 0 when i > j.
(b) When i = j, the only nonzero term in the product Ai∗B∗i is aiibii.
(c) Yes.
3.5.9. Use [AB]ij= kaikbkj along with the rules of differentiation to write
d[AB]ij
dt =
d( kaikbkj)
dt =
k
d(aikbkj)
dt
=
k
daik
dt bkj+aik dbkj
dt
=
k
daik
dt bkj+
k
aikdbkj
dt
=
dA
dt B
ij
+
AdB
dt
ij
=
dA
dt B+A dB
dt
ij
.
3.5.10. (a) [Ce]i= the total number of pathsleaving node i.
(b) [eTC]
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3.5.11. At time t, the concentration of salt in tank i is xi(t)
V lbs/gal. For tank 1,
dx1
dt =
lbs
sec coming in− lbs
sec going out = 0 lbs sec−
rgal
sec×
x1(t)
V
lbs gal
=−Vrx1(t)lbs
sec.
For tank 2,
dx2
dt =
lbs
sec coming in− lbs
sec going out =
r Vx1(t)
lbs sec−
rgal
sec ×
x2(t)
V
lbs gal
= r
Vx1(t)
lbs sec−
r Vx2(t)
lbs sec =
r V
x1(t)−x2(t)
,
and for tank 3,
dx3
dt =
lbs
sec coming in− lbs
sec going out =
r Vx2(t)
lbs sec−
rgal
sec ×
x3(t)
V
lbs gal
= r
Vx2(t)
lbs sec−
r Vx3(t)
lbs sec =
r V
x2(t)−x3(t)
.
This is a system of three linear first-order differential equations
dx1
dt = r
V
−x1(t)
dx2
dt = r
V
x1(t) −x2(t)
dx3
dt = r
V
x2(t) −x3(t)
that can be written as a single matrix differential equation
dx1/dt
dx2/dt
dx3/dt
= r
V
−1 0 0
1 −1 0
0 1 −1
x1(t)
x2(t)
x3(t)
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Solutions for exercises in section 3. 6
3.6.1.
AB=
A11 A12 A13
A21 A22 A23
BB12
B3
=
A11B1+A12B2+A13B3
A21B1+A22B2+A23B3
=
−10 −19
−10 −19
−1 −1
3.6.2. Use block multiplication to verify L2 =I—be careful not to commute any of
the terms when forming the various products.
3.6.3. Partition the matrix as A=
I C
0 C
, where C= 13
11 11 11
1 1 1
and observe
that C2=C. Use this together with block multiplication to conclude that
Ak=
I C+C2+C3+· · ·+Ck
0 Ck
=
I kC
0 C
.
Therefore, A300=
1 0 0 100 100 100
0 1 0 100 100 100
0 0 1 100 100 100
0 0 0 1/3 1/3 1/3 0 0 0 1/3 1/3 1/3 0 0 0 1/3 1/3 1/3
.
3.6.4. (A∗A)∗=A∗A∗∗=A∗A and (AA∗)∗=A∗∗A∗=AA∗.
3.6.5. (AB)T =BTAT =BA=AB. It is easy to construct a 2×2 example to show
that this need not be true when AB=BA.
3.6.6.
[(D+E)F]ij = (D+E)i∗F∗j =
k
[D+E]ik[F]kj=
k
([D]ik+ [E]ik) [F]kj
=
k
([D]ik[F]kj+ [E]ik[F]kj) =
k
[D]ik[F]kj+
k
[E]ik[F]kj
=Di∗F∗j+Ei∗F∗j= [DF]ij+ [EF]ij
= [DF+EF]ij.
3.6.7. If a matrix X did indeed exist, then
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which is impossible.
3.6.8. (a) yTA=bT =
⇒ (yTA)T =bT T =
⇒ ATy =b. This is an n
×m
system of equations whose coefficient matrix is AT. (b) They are the same.
3.6.9. Draw a transition diagram similar to that in Example 3.6.3 with North and South
replaced by ON and OFF, respectively. Let xk be the proportion of switches in
the ON state, and let yk be the proportion of switches in the OFF state after
k clock cycles have elapsed. According to the given information,
xk =xk−1(.1) +yk−1(.3)
yk =xk−1(.9) +yk−1(.7)
so that pk=pk−1P, where
pk= (xk yk) and P=
.1 .9
.3 .7
.
Just as in Example 3.6.3, pk =p0Pk. Compute a few powers of P to find
P2=
.280 .720
.240 .760
, P3=
.244 .756
.252 .748
P4=
.251 .749
.250 .750
, P5=
.250 .750
.250 .750
and deduce that P∞= lim
k→∞Pk =
1/4 3/4 1/4 3/4
. Thus
pk→p0P∞= (14(x0+y0) 34(x0+y0) ) = (14 34).
For practical purposes, the device can be considered to be in equilibrium after about 5 clock cycles—regardless of the initial proportions.
3.6.10. (−4 1 −6 5 )
3.6.11. (a) trace(ABC) = trace(A{BC}) = trace({BC}A) = trace(BCA). The
other equality is similar. (b) Use almost any set of 2×2 matrices to con-struct an example that shows equality need not hold. (c) Use the fact that
traceCT=trace(C) for all square matrices to conclude that
traceATB=trace(ATB)T=traceBTAT T
=traceBTA=traceABT.
3.6.12. (a) xTx= 0
⇐⇒ nk=1x2i = 0⇐⇒xi= 0 for each i⇐⇒x=0.
(b) traceATA= 0
⇐⇒
i
[ATA]
ii= 0⇐⇒
i
(AT)
i∗A∗i= 0
⇐⇒
i
k
[AT]ik[A]ki= 0⇐⇒
i
k
[A]ki[A]ki= 0
⇐⇒
i
k
[A]2
ki= 0
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Solutions for exercises in section 3. 7
3.7.1. (a)
3 −2
−1 1
(b) Singular (c)
24 −−47 34 5 −8 4
(d) Singular
(e)
2 −1 0 0
−1 2 −1 0
0 −1 2 −1
0 0 −1 1
3.7.2. Write the equation as (I−A)X=B and compute
X= (I−A)−1B=
10 −11 −11
0 0 1
12 21
3 3 =
−21 −42
3 3
.
3.7.3. In each case, the given information implies that rank(A)< n—see the solution
for Exercise 2.1.3.
3.7.4. (a) If D is diagonal, then D−1 exists if and only if each dii= 0, in which case
d11 0 · · · 0
0 d22 · · · 0
..
. ... . .. ... 0 0 · · · dnn
−1
=
1/d11 0 · · · 0
0 1/d22 · · · 0
..
. ... . .. ... 0 0 · · · 1/dnn
.
(b) If T is triangular, then T−1 exists if and only if each t
ii = 0. If T
is upper (lower) triangular, then T−1 is also upper (lower) triangular with
[T−1]
ii= 1/tii.
3.7.5. A−1T =AT−1=A−1.
3.7.6. Start with A(I−A) = (I−A)A and apply (I−A)−1 to both sides, first on
one side and then on the other.
3.7.7. Use the result of Example 3.6.5 that says that trace(AB) = trace(BA) to
write
m=trace(Im) =trace(AB) =trace(BA) =trace(In) =n.
3.7.8. Use the reverse order law for inversion to write
A(A+B)−1B−1=B−1(A+B)A−1=B−1+A−1
and
B(A+B)−1A−1=A−1(A+B)B−1=B−1+A−1.
3.7.9. (a) (I−S)x=0 =⇒ xT(I−S)x= 0 =⇒ xTx=xTSx. Taking
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(recall Exercise 3.6.12). The conclusion follows from property (3.7.8).
(b) First notice that Exercise 3.7.6 implies that A = (I+S)(I−S)−1 =
(I−S)−1(I+S). By using the reverse order laws, transposing both sides yields
exactly the same thing as inverting both sides.
3.7.10. Use block multiplication to verify that the product of the matrix with its inverse
is the identity matrix.
3.7.11. Use block multiplication to verify that the product of the matrix with its inverse
is the identity matrix.
3.7.12. Let M=
A B
C D
and X=
DT −BT
−CT AT
. The hypothesis implies that
MX = I, and hence (from the discussion in Example 3.7.2) it must also be true that XM = I, from which the conclusion follows. Note: This problem appeared on a past Putnam Exam—a national mathematics competition for undergraduate students that is considered to be quite challenging. This means that you can be proud of yourself if you solved it before looking at this solution.
Solutions for exercises in section 3. 8
3.8.1. (a) B−1=
10 −21 −11
1 4 −2
(b) Let c= 0 0 1
and dT = ( 0 2 1 ) to obtain C−1=
0 −2 1
1 3 −1
−1 −4 2
3.8.2. A∗j needs to be removed, and b needs to be inserted in its place. This is
accomplished by writing B=A+(b−A∗j)eTj. Applying the Sherman–Morrison
formula with c=b−A∗j and dT =eTj yields
B−1=A−1−A
−1(b
−A∗j)eTjA−1
1 +eT
jA−1(b−A∗j)
=A−1−A
−1beT
jA−1−ejeTjA−1
1 +eT
jA−1b−eTjej
=A−1−A−
1b[A−1]
j∗−ej[A−1]j∗
[A−1]
j∗b =A
−1
−
A−1b−e
j[A−1]j∗
[A−1]
j∗b .
3.8.3. Use the Sherman–Morrison formula to write
z= (A+cdT)−1b=
A−1−A
−1cdTA−1
1 +dTA−1c
b=A−1b−A
−1cdTA−1b
1 +dTA−1c
=x− yd
Tx
1 +dTy.
3.8.4. (a) For a nonsingular matrix A, the Sherman–Morrison formula guarantees
that A+αeieTj is also nonsingular when 1 +α
A−1
ji= 0, and this certainly
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(b) Write Em×m= [ǫij] = mi,j=1ǫijeieTj and successively apply part (a) to
I+E=
I+ǫ11e1eT1
+ǫ12e1eT2
+· · ·+ǫmmemeTm
to conclude that when the ǫij’s are sufficiently small,
I+ǫ11e1eT1,
I+ǫ11e1eT1
+ǫ12e1eT2
, . . . , I+E
are each nonsingular.
3.8.5. Write A+ǫB=A(I+A−1ǫB). You can either use the Neumann series result
(3.8.5) or Exercise 3.8.4 to conclude that (I+A−1ǫB) is nonsingular whenever the entries of A−1ǫB are sufficiently small in magnitude, and this can be insured
by restricting ǫ to a small enough interval about the origin. Since the product of two nonsingular matrices is again nonsingular—see (3.7.14)—it follows that
A+ǫB=A(I+A−1ǫB) must be nonsingular.
3.8.6. Since
I C
0 I
A C
DT
−I
I 0
DT I
=
A+CDT 0
0 −I
,
we can use R=DT and B=
−I in part (a) of Exercise 3.7.11 to obtain
I 0
−DT I
A−1+A−1CS−1DTA−1
−A−1CS−1
−S−1DTA−1 S−1
I −C
0 I
=
A+CDT−1 0
0 −I
,
where S=−I+DTA−1C. Comparing the upper-left-hand blocks produces
A+CDT−1=A−1−A−1CI+DTA−1C−1DTA−1.
3.8.7. The ranking from best to worst condition is A, B, C, because
A−1= 1
100
2 1 11 2 1 1 1 1
=⇒ κ(A) = 20 = 2×101
B−1=
−
1465 −161 17
173 19 −2
−82 −9 1
=⇒ κ(B) = 149,513≈1.5×105
C−1=
−426592025 −397941889 −94845
45 −42 1
=⇒ κ(C) = 82,900,594≈8.2×107.
3.8.8. (a) Differentiate A(t)A(t)−1=I with the product rule for differentiation
(re-call Exercise 3.5.9).
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Solutions for exercises in section 3. 9
3.9.1. (a) If G1,G2, . . . ,Gk is the sequence of elementary matrices that corresponds
to the elementary row operations used in the reduction [A|I]−→[B|P], then
Gk· · ·G2G1[A|I] = [B|P] =⇒ [Gk· · ·G2G1A|Gk· · ·G2G1I] = [B|P]
=⇒ Gk· · ·G2G1A=B and Gk· · ·G2G1=P.
(b) Use the same argument given above, but apply it on the right-hand side. (c) [A|I]−−−−−−−−→Gauss–Jordan [EA|P] yields
12 24 36 47 10 01 00
1 2 3 6 0 0 1
−→
10 20 30 01 −72 −14 00
0 0 0 0 −5 2 1
.
Thus P =
−72 −4 01 0
−5 2 1
is the product of the elementary matrices
corre-sponding to the operations used in the reduction, and PA=EA.
(d) You already have P such that PA=EA. Now find Q such that EAQ=
Nr by column reducing EA. Proceed using part (b) to accumulate Q.
EA
I4
−→
1 2 3 0 0 0 0 1 0 0 0 0
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
−→
1 0 2 3 0 1 0 0 0 0 0 0
1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0
−→
1 0 0 0
0 1 0 0
0 0 0 0
1 0 −2 −3
0 0 1 0
0 0 0 1
0 1 0 0
3.9.2. (a) Yes—because rank(A) = rank(B). (b) Yes—because EA = EB.
(c) No—because EAT =EBT.
3.9.3. The positions of the basic columns in A correspond to those in EA. Because
A row∼ B⇐⇒ EA =EB, it follows that the basic columns in A and B must
be in the same positions.
3.9.4. An elementary interchange matrix (a Type I matrix) has the form E=I−uuT,
where u = ei −ej, and it follows from (3.9.1) that E = ET = E−1. If
P=E1E2· · ·Ek is a product of elementary interchange matrices, then the
re-verse order laws yield
P−1= (E1E2· · ·Ek)−1=E−k1· · ·E−21E−11
=ET
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3.9.5. They are all true! A∼I∼A−1 because rank(A) =n=rankA−1, Arow
∼
A−1 because PA=A−1 with P=A−12=A−2, and Acol
∼A−1 because
AQ =A−1 with Q=A−2. The fact that Arow
∼ I and Acol∼ I follows since
A−1A=AA−1=I.
3.9.6. (a), (c), (d), and (e) are true.
3.9.7. Rows i and j can be interchanged with the following sequence of Type II and
Type III operations—this is Exercise 1.2.12 on p. 14.
Rj←Rj+Ri (replace rowj by the sum of rowj andi)
Ri←Ri−Rj (replace rowi by the difference of rowiandj)
Rj←Rj+Ri (replace rowj by the sum of rowj andi)
Ri← −Ri (replace rowiby its negative)
Translating these to elementary matrices (remembering to build from the right to the left) produces
(I−2eieTi)(I+ejeTi)(I−eiejT)(I+ejeTi) =I−uuT, where u=ei−ej.
3.9.8. Let Bm×r= [A∗b1A∗b2· · ·A∗br] contain the basic columns of A, and let Cr×n
contain the nonzero rows of EA. If A∗k is basic—say A∗k = A∗bj—then
C∗k=ej, and
(BC)∗k=BC∗k=Bej =B∗j=A∗bj =A∗k.
If A∗k is nonbasic, then C∗k is nonbasic and has the form
C∗k=
µ1 µ2 .. . µj .. . 0
=µ1
1 0 .. . 0 .. . 0
+µ2
0 1 .. . 0 .. . 0
+· · ·+µj
0 0 .. . 1 .. . 0
=µ1e1+µ2e2+· · ·+µjej,
where the ei’s are the basic columns to the left of C∗k. Because Arow∼ EA,
the relationships that exist among the columns of A are exactly the same as the relationships that exist among the columns of EA. In particular,
A∗k=µ1A∗b1+µ2A∗b2+· · ·+µjA∗bj,
where the A∗bi’s are the basic columns to the left of A∗k. Therefore,
(BC)∗k=BC∗k=B(µ1e1+µ2e2+· · ·+µjej)
=µ1B∗1+µ2B∗2+· · ·+µjB∗j
=µ1A∗b1+µ2A∗b2+· · ·+µjA∗bj
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3.9.9. If A=uvT, where u
m×1 and vn×1 are nonzero columns, then
urow∼ e1 and vT
col
∼eT
1 =⇒ A=uvT ∼e1eT1 =N1 =⇒ rank(A) = 1.
Conversely, if rank(A) = 1, then the existence of u and v follows from Exer-cise 3.9.8. If you do not wish to rely on ExerExer-cise 3.9.8, write PAQ=N1=e1eT1,
where e1 is m×1 and eT1 is 1×n so that
A=P−1e1eT1Q−1=
P−1∗1Q−11∗=uvT.
3.9.10. Use Exercise 3.9.9 and write
A=uvT =⇒ A2=uvT uvT=uvTuvT =τuvT =τA,
where τ = vTu. Recall from Example 3.6.5 that trace(AB) = trace(BA),
and write
τ=trace(τ) =tracevTu=traceuvT=trace(A).
Solutions for exercises in section 3. 10
3.10.1. (a) L =
1 0 0 4 1 0 3 2 1
and U =
1 4 5 0 2 6 0 0 3
(b) x1 =
110
−36 8
and
x2=
−11239
10
(c) A−1=1 6
−12442 −4015 −146
10 −4 2
3.10.2. (a) The second pivot is zero. (b) P is the permutation matrix associated
with the permutation p= ( 2 4 1 3 ). P is constructed by permuting the rows of I in this manner.
L=
1 0 0 0
0 1 0 0
1/3 0 1 0
2/3 −1/2 1/2 1
and U=
3 6 −12 3
0 2 −2 6
0 0 8 16
0 0 0 −5
(c) x=
2
−1 0 1
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3.10.3. ξ= 0, ±√2, ±√3
3.10.4. A possesses an LU factorization if and only if all leading principal submatrices
are nonsingular. The argument associated with equation (3.10.13) proves that
Lk 0
cTU−1
k 1
Uk L−k1b
0 ak+1,k+1−cTA−k1b
=Lk+1Uk+1
is the LU factorization for Ak+1. The desired conclusion follows from the fact
that the k+ 1th pivot is the (k+ 1, k+ 1) -entry in U
k+1. This pivot must be
nonzero because Uk+1 is nonsingular.
3.10.5. If L and U are both triangular with 1’s on the diagonal, then L−1 and U−1
contain only integer entries, and consequently A−1 = U−1L−1 is an integer
matrix.
3.10.6. (b) L=
1 0 0 0
−1/2 1 0 0
0 −2/3 1 0
0 0 −3/4 1
and U=
2 −1 0 0
0 3/2 −1 0 0 0 4/3 −1
0 0 0 1/4
3.10.7. Observe how the LU factors evolve from Gaussian elimination. Following the
procedure described in Example 3.10.1 where multipliers ℓij are stored in the
positions they annihilate (i.e., in the (i, j) -position), and where ⋆’s are put in positions that can be nonzero, the reduction of a 5×5 band matrix with bandwidth w= 2 proceeds as shown below.
⋆ ⋆ ⋆ 0 0
⋆ ⋆ ⋆ ⋆ 0
⋆ ⋆ ⋆ ⋆ ⋆
0 ⋆ ⋆ ⋆ ⋆
0 0 ⋆ ⋆ ⋆
−→
⋆ ⋆ ⋆ 0 0
l21 ⋆ ⋆ ⋆ 0
l31 ⋆ ⋆ ⋆ ⋆
0 ⋆ ⋆ ⋆ ⋆
0 0 ⋆ ⋆ ⋆
−→
⋆ ⋆ ⋆ 0 0
l21 ⋆ ⋆ ⋆ 0
l31 l32 ⋆ ⋆ ⋆
0 l42 ⋆ ⋆ ⋆
0 0 ⋆ ⋆ ⋆
−→
⋆ ⋆ ⋆ 0 0
l21 ⋆ ⋆ ⋆ 0
l31 l32 ⋆ ⋆ ⋆
0 l42 l43 ⋆ ⋆
0 0 l53 ⋆ ⋆
−→
⋆ ⋆ ⋆ 0 0
l21 ⋆ ⋆ ⋆ 0
l31 l32 ⋆ ⋆ ⋆
0 l42 l43 ⋆ ⋆
0 0 l53 l54 ⋆
Thus L=
1 0 0 0 0
l21 1 0 0 0
l31 l32 1 0 0
0 l42 l43 1 0
0 0 l53 l54 1
and U=
⋆ ⋆ ⋆ 0 0 0 ⋆ ⋆ ⋆ 0 0 0 ⋆ ⋆ ⋆
0 0 0 ⋆ ⋆
0 0 0 0 ⋆
.
3.10.8. (a) A=
0 1 1 0
(b) A=
1 0
0 −1
3.10.9. (a) L=
1 0 04 1 0 3 2 1
, D=
1 0 00 2 0 0 0 3
, and U=
1 4 50 1 3 0 0 1
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(b) Use the same argument given for the uniqueness of the LU factorization with minor modifications.
(c) A = AT =
⇒ LDU = UTDTLT = UTDLT. These are each LDU
factorizations for A, and consequently the uniqueness of the LDU factorization means that U=LT.
3.10.10. A is symmetric with pivots 1, 4, 9. The Cholesky factor is R=
1 0 0
2 2 0
3 3 3
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Solutions for Chapter 4
Solutions for exercises in section 4. 1
4.1.1. Only (b) and (d) are subspaces.
4.1.2. (a), (b), (f), (g), and (i) are subspaces.
4.1.3. All of ℜ3.
4.1.4. If v∈ V is a nonzero vector in a space V, then all scalar multiples αv must
also be in V.
4.1.5. (a) A line. (b) The (x,y)-plane. (c) ℜ3
4.1.6. Only (c) and (e) span ℜ3. To see that (d) does not span ℜ3, ask whether
or not every vector (x, y, z) ∈ ℜ3 can be written as a linear combination of
the vectors in (d). It’s convenient to think in terms columns, so rephrase the
question by asking if every b=
x y z
can be written as a linear combination
of v1=
12
1 , v2=
20
−1 , v3=
44
1
. That is, for each b∈ ℜ
3, are
there scalars α1, α2, α3 such that α1v1+α2v2+α3v3=b or, equivalently, is
12 2 40 4 1 −1 1
αα12
α3
=
xy
z
consistent for all xy
z
?
This is a system of the form Ax=b, and it is consistent for all b if and only if rank([A|b]) =rank(A) for all b. Since
12 2 40 4 xy
1 −1 1 z
→
10 −24 −44 y−x2x
0 −3 −3 z−x
→
10 −24 −44 y−x2x
0 0 0 (x/2)−(3y/4) +z
,
it’s clear that there exist b’s (e.g., b= (1,0,0)T) for which Ax =b is not
consistent, and hence not all b’s are a combination of the vi’s. Therefore, the
vi’s don’t span ℜ3.
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4.1.8. (a) u,v ∈ X ∩ Y =⇒ u,v ∈ X and u,v ∈ Y. Because X and Y are
closed with respect to addition, it follows that u+v ∈ X and u+v ∈ Y,
and therefore u+v∈ X ∩ Y. Because X and Y are both closed with respect to scalar multiplication, we have that αu ∈ X and αu ∈ Y for all α, and consequently αu∈ X ∩ Y for all α.
(b) The union of two different lines through the origin in ℜ2 is not a subspace.
4.1.9. (a) (A1) holds because x1,x2 ∈A(S) =⇒ x1 =As1 and x2 =As2 for
some s1,s2 ∈ S =⇒ x1+x2 = A(s1+s2). Since S is a subspace, it is
closed under vector addition, so s1+s2∈ S. Therefore, x1+x2 is the image of
something in S—namely, s1+s2—and this means that x1+x2∈A(S). To see
that (M1) holds, consider αx, where α is an arbitrary scalar and x∈A(S).
Now, x ∈A(S) =⇒ x=As for some s∈ S =⇒ αx=αAs=A(αs).
Since S is a subspace, we are guaranteed that αs∈ S, and therefore αx is the image of something in S. This is what it means to say αx∈A(S).
(b) Prove equality by demonstrating that span{As1,As2, . . . ,Ask} ⊆ A(S)
and A(S)⊆span{As1,As2, . . . ,Ask}. To show span{As1,As2, . . . ,Ask} ⊆
A(S), write
x∈span{As1,As2, . . . ,Ask} =⇒ x= k
i=1
αi(Asi) =A
k
i=1
αisi
∈A(S).
Inclusion in the reverse direction is established by saying
x∈A(S) =⇒ x=Asfor some s∈ S =⇒ s=
k
i=1
βisi
=⇒ x=A
k
i=1
βisi
=
k
i=1
βi(Asi)∈span{As1,As2, . . . ,Ask}.
4.1.10. (a) Yes, all of the defining properties are satisfied.
(b) Yes, this is essentially ℜ2.
(c) No, it is not closed with respect to scalar multiplication.
4.1.11. If span(M) =span(N), then every vector in N must be a linear combination
of vectors from M. In particular, v must be a linear combination of the mi’s,
and hence v∈span(M). To prove the converse, first notice that span(M)⊆
span(N). The desired conclusion will follow if it can be demonstrated that
span(M) ⊇ span(N). The hypothesis that v ∈ span(M) guarantees that
v= ri=1βimi. If z∈span(N), then
z=
r
i=1
αimi+αr+1v=
r
i=1
αimi+αr+1
r
i=1
βimi
=
r
i=1
αi+αr+1βi
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which shows z∈span(M), and therefore span(M)⊇span(N).
4.1.12. To show span(S) ⊆ M, observe that x ∈ span(S) =⇒ x = iαivi.
If V is any subspace containing S, then iαivi ∈ V because V is closed
under addition and scalar multiplication, and therefore x∈ M. The fact that
M ⊆span(S) follows because if x∈ M, then x∈span(S) because span(S) is one particular subspace that contains S.
Solutions for exercises in section 4. 2
4.2.1. R(A) =span
1 −2 1 , 1 0 2 , N
AT=span
4 1 −2 ,
N(A) =span
−2 1 0 0 0 , 2 0 −3 1 0 , −1 0 −4 0 1 ,
RAT=span
1 2 0 −2 1 , 0 0 1 3 4 .
4.2.2. (a) This is simply a restatement of equation (4.2.3).
(b) Ax=b has a unique solution if and only if rank(A) =n (i.e., there are no free variables—see§2.5), and (4.2.10) says rank(A) =n⇐⇒N(A) ={0}.
4.2.3. (a) It is consistent because b∈R(A).
(b) It is nonunique because N(A)={0}—see Exercise 4.2.2.
4.2.4. Yes, because rank[A|b] = rank(A) = 3 =⇒ ∃ xsuch thatAx =b—i.e.,
Ax=b is consistent.
4.2.5. (a) If R(A) =ℜn, then
R(A) =R(In) =⇒ A
col
∼ In =⇒ rank(A) =rank(In) =n.
(b) R(A) =RAT=
ℜn and N(A) =NAT=
{0}.
4.2.6. EA = EB means that R
AT = RBT and N(A) = N(B). However,
EAT =EBT implies that R(A) =R(B) and N
AT=NBT.
4.2.7. Demonstrate that rank(An×n) =n by using (4.2.10). If x∈N(A), then
Ax=0 =⇒ A1x=0 and A2x=0
=⇒ x∈N(A1) =RAT2
=⇒ ∃ yT such that xT =yTA2
=⇒ xTx=yTA2x=0 =⇒
i
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4.2.8. yTb= 0∀y∈NAT=RPT
2
=⇒ P2b= 0 =⇒ b∈N(P2) =R(A)
4.2.9. x∈ RA | B ⇐⇒ ∃ y such thatx= A| By =A | B
y1
y2
=Ay1+
By2⇐⇒x∈R(A) +R(B)
4.2.10. (a) p+N(A) is the set of all possible solutions to Ax=b. Recall from (2.5.7)
that the general solution of a nonhomogeneous equation is a particular solution plus the general solution of the homogeneous equation Ax = 0. The general solution of the homogeneous equation is simply a way of describing all possible solutions of Ax=0, which is N(A).
(b) rank(A3×3) = 1 means that N(A) is spanned by two vectors, and hence
N(A) is a plane through the origin. From the parallelogram law, p+N(A) is a plane parallel to N(A) passing through the point defined by p.
(c) This time N(A) is spanned by a single vector, and p+N(A) is a line parallel to N(A) passing through the point defined by p.
4.2.11. a∈RAT
⇐⇒ ∃y such that aT =yTA. If Ax=b, then
aTx=yTAx=yTb,
which is independent of x.
4.2.12. (a) b∈R(AB) =⇒ ∃x such that b=ABx=A(Bx) =⇒ b∈R(A)
because b is the image of Bx.
(b) x∈N(B) =⇒ Bx=0 =⇒ ABx=0 =⇒ x∈N(AB).
4.2.13. Given any z ∈R(AB), the object is to show that z can be written as some
linear combination of the Abi’s. Argue as follows. z∈R(AB) =⇒ z=ABy
for some y. But it is always true that By∈R(B), so
By=α1b1+α2b2+· · ·+αnbn,
and therefore z=ABy=α1Ab1+α2Ab2+· · ·+αnAbn.
Solutions for exercises in section 4. 3
4.3.1. (a) and (b) are linearly dependent—all others are linearly independent. To write
one vector as a combination of others in a dependent set, place the vectors as columns in A and find EA. This reveals the dependence relationships among
columns of A.
4.3.2. (a) According to (4.3.12), the basic columns in A always constitute one maximal
linearly independent subset.
(b) Ten—5 sets using two vectors, 4 sets using one vector, and the empty set.
4.3.3. rank(H)≤3, and according to (4.3.11), rank(H) is the maximal number of
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4.3.4. The question is really whether or not the columns in
ˆ A=
S L F
#1 1 1 1 10
#2 1 2 1 12
#3 1 2 2 15
#4 1 3 2 17
are linearly independent. Reducing Aˆ to EAˆ shows that 5 + 2S+ 3L−F = 0.
4.3.5. (a) This follows directly from the definition of linear dependence because there
are nonzero values of α such that α0=0.
(b) This is a consequence of (4.3.13).
4.3.6. If each tii = 0, then T is nonsingular, and the result follows from (4.3.6) and
(4.3.7).
4.3.7. It is linearly independent because
α1
1 0
0 0
+α2
1 1 0 0
+α3
1 1 1 0
+α4
1 1 1 1 = 0 0 0 0
⇐⇒ α1
1 0 0 0 +α2
1 1 0 0 +α3
1 1 1 0 +α4
1 1 1 1 = 0 0 0 0 ⇐⇒
1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 1
α1 α2 α3 α4 = 0 0 0 0 ⇐⇒ α1 α2 α3 α4 = 0 0 0 0 .
4.3.8. A is nonsingular because it is diagonally dominant.
4.3.9. S is linearly independent using exact arithmetic, but using 3-digit arithmetic
yields the conclusion that S is dependent.
4.3.10. If e is the column vector of all 1’s, then Ae=0, so that N(A)={0}.
4.3.11. (Solution 1.) iαiPui = 0 =⇒ P iαiui = 0 =⇒ iαiui =
0 =⇒ eachαi = 0 because the ui’s are linearly independent.
(Solution 2.) If Am×n is the matrix containing the ui’s as columns, then
PA=B is the matrix containing the vectors in P(S) as its columns. Now,
Arow∼ B =⇒ rank(B) =rank(A) =n,
and hence (4.3.3) insures that the columns of B are linearly independent. The result need not be true if P is singular—take P=0 for example.
4.3.12. If Am×n is the matrix containing the ui’s as columns, and if
Qn×n=
1 1 · · · 1 0 1 · · · 1 ..
. ... . .. ... 0 0 · · · 1
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then the columns of B=AQ are the vectors in S′. Clearly, Q is nonsingular
so that Acol∼ B, and thus rank(A) =rank(B). The desired result now follows from (4.3.3).
4.3.13. (a) and (b) are linearly independent because the Wronski matrix W(0) is
non-singular in each case. (c) is dependent because sin2x−cos2x+ cos 2x= 0.
4.3.14. If S were dependent, then there would exist a constant α such that x3=α|x|3
for all values of x. But this would mean that
α= x
3
|x|3 =
1 ifx >0, −1 ifx <0,
which is clearly impossible since α must be constant. The associated Wronski matrix is
W(x) =
x3 x3
3x2 3x2
whenx≥0,
x3
−x3
3x2
−3x2
whenx <0,
which is singular for all values of x.
4.3.15. Start with the fact that
AT diag. dom. =⇒ |bii|>|di|+
j=i
|bji| and |α|>
j
|cj|
=⇒
j=i
|bji|<|bii| − |di| and
1
|α|
j=i
|cj|<1−|
ci|
|α|,
and then use the forward and backward triangle inequality to write
j=i
|xij|=
j=i
bji−dicj
α
≤
j=i
|bji|+|di|
|α|
j=i
|cj|
<|bii| − |di|+|di|
1−|ci|
|α|
=|bii| −|di| |ci|
|α| ≤
bii−dici
α
=|xii|.
Now, diagonal dominance of AT insures that α is the entry of largest
